RegEx用于传递除字符列表以外的所有字符

Question

I want a pattern that contains all characters except English and Persian numbers. I found this pattern but my problem is how can I do against this pattern.

For example contains * and / and ... but not contains 1 2 3 ۱ ۲ ۳.

This pattern gets all numbers:

^[\u0600-\u06FF\s0-9]+$

Answer 1

We might be able to solve this problem by adding our undesired unicodes in a char list, then swiping everything else, which I'm unsure if space would be undesired or not.

Maybe, something like this with modification would work:

([\s\S].*?)([\x{0600}-\x{06FF}0-9]+)?

Demo

We would just replace x with u for JavaScript:

 const regex = /([\\s\\S].*?)([\\u{0600}-\\u{06FF}0-9]+)?/gmu; const str = `everthing we wish to have before 1 2 3 ۱ ۲ ۳ and everything else we wish to have`; const subst = `$1`; // The substituted value will be contained in the result variable const result = str.replace(regex, subst); console.log('Substitution result: ', result); 

RegEx

If this wasn't your desired expression, you can modify/change your expressions in regex101.com .

Answer 2

这不是ASCII，也不是波斯数字

[^0-9\۰-\۹]+

Answer 3

This should works

/[^\d۱۲۳۴۵۶۷۸۹]/g

Demo

Note: If I forget some Persian digits just add it before ]